Skip to main content

Algebra-II: Companion notes

Abhay Soman

Section 9.4 Submodules

Definition 9.4.1.

Let \(M\) be a left \(R\)-module. A subgroup \(N\) of \(M\) is said to be a submodule of \(R\) or \(R\)-submodule if \(N\) is closed under the ‘scalar multiplication’ of \(R\) on \(M\text{,}\) i.e., for any \(r\in R\) and any \(n\in N\) we have \(r\cdot n\in N\text{.}\)

Convention 9.4.2.

A submodule \(N\) of a module \(M\) is denoted by \(N\leq M\text{.}\)

Definition 9.4.3. (Submodule generated by a subset).

Let \(M\) be a left \(R\)-module, and let \(X\) be a subset of \(M\text{.}\) The \(R\)-submodule generated by \(X\) is
\begin{equation*} \langle X\rangle=\left\{\sum_{i=1}^nr_ix_i:x_i\in M,r_i\in R\;\text{and}\;n\in\N\right\}. \end{equation*}
If \(X\) is a finite set, say \(X=\{x_1,\ldots,x_n\}\) then we write the \(R\)-module generated by \(X\) as
\begin{equation*} Rx_1+Rx_2+\cdots+Rx_n \end{equation*}
and we call such an \(R\)-module a finitely generated \(R\)-module.
In particular, if \(X={x}\) then the finitely generated \(R\)-module is called principal \(R\)-module or cyclic \(R\)-module.

Definition 9.4.4.

Let \(M\) be a left \(R\)-module. An element \(m\in M\) is said to be an \(R\)-linear combination of \(x_1,\ldots,x_n\in M\) if
\begin{equation*} m=r\cdot x_1+\cdots+r\cdot x_n. \end{equation*}
Let \(R\) be a ring and let \(M\) be a left \(R\)-module. The trivial subgroup \(\{0\}\leq M\) is an \(R\)-submodule of \(M\text{.}\) Also, module \(M\) is an \(R\)-submodule of itself.
If \(V\) is an \(F\)-vector space then any subspace of \(V\) is an \(F\)-submodule of \(V\text{.}\)
Every subgroup of an abelian group is a \(\Z\)-submodule, refer to Example 9.2.1.
Consider a ring \(R\) as a left module over itself via the left multiplication (see Example 9.2.5). If \(I\) is a left ideal of \(R\) then \(I\) is a left \(R\)-module.
Let \(M\) be a left \(R\)-module. By Example 9.2.6 there is a ring homomorphism \(R\to\End_{\rm Grps}(M)\text{.}\) The kernel of this ring homomorphism is called the annihilator of \(M\):
\begin{equation*} {\rm Ann}_R(M)=\{r\in R:r\cdot m=0\;\text{for all}\;m\in M\} \end{equation*}
being a two-sided ideal of \(R\) is a left \(R\)-submodule of \(R\text{.}\)
Consider \(\Z/n\Z\) as a \(\Z\)-module. In this case \(\Ann_{\Z}(\Z/n\Z)=n\Z\text{.}\)
Let \(V\) be a vector space over a field \(F\text{,}\) i.e., \(V\) is an \(F\)-mod. Fix a linear transformation \(T\in\End_F(V)\text{.}\) As in Example 9.2.4 we consider \(V\) as an \(F[X]\)-mod. We show that submodules of \(V\) as an \(F[X]\)-mod are precisely \(T\)-invariant \(F\)-subspaces of a vector space \(V\text{.}\)
Assume that \(W\) is an \(F[X]\)-submodule of \(V\text{.}\) First observe that \(W\) is also an \(F\)-subspace of \(V\text{.}\) Indeed, under \(F[X]\)-module structure any constant polynomial \(a\in F[X]\) acts on \(w\in W\) by the following way:
\begin{equation*} (a,w)\mapsto a\unit_V(w)=a\cdot w\in W, \end{equation*}
where \(a\cdot w\) is the scalar multiplication via the vector space structure. For any \(w\in W\) we have \((X,w)\mapsto T(w)\in W\) and hence \(W\) is a \(T\)-invariant subspace of \(V\text{.}\)
Conversely assume that \(W\) is a \(T\)-invariant subspace of \(V\text{.}\) Then for any \(n\in\N\) and any \(a_n\in F\) we have \(W\) is \(a_nT^n\)-invariant subspace of \(V\text{.}\) Therefore, for any \(w\in W\) and any \(a_0+a_X+\cdots+a_nX^n\in F[X]\text{,}\) we get that
\begin{equation*} \left(a_0+a_1X+\cdots+a_nT^n,w\right)\mapsto a_0\unit_V(w)+a_1T(w)+\cdots+a_nT^n(w)\in W \end{equation*}
Hence \(W\) is an \(F[X]\)-submodule.

Definition 9.4.11. (Proper submodule and non-trivial submodule).

A left \(R\)-submodule \(N\leq M\) is said to be proper if \(N\neq M\) and \(N\) is said to be non-trivial if \(N\neq\{0\}\text{.}\)

Definition 9.4.12. (Simple module).

Let \(R\) be a ring and let \(M\) be an \(R\)-module. We say that \(M\) is simple module if \(M\) has no proper non-trivial submodule.
Let \(0\neq m\in M\text{.}\) Consider the left non-trivial submodule generated by \(m\text{,}\) \(Rm\text{.}\) Since \(M\) is simple and \(Rm\neq \{0\}\) we have \(M=Rm\text{.}\)
In Example 9.2.1 we showed that \(\Z\)-modules are precisely abelian groups. Thus a simple \(\Z\)-module must be a cyclic group of a prime order.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) As in Example 9.2.3 we consider \(V\) as an \(\End_F(V)\)-module. We show that \(V\) is a simple \(\End_F(V)\)-module.
In view of Lemma 9.4.13 it is enough to show that given any \(v,w\in V\) with \(v\neq 0\) there exists \(T\in\End_F(V)\) such that \(T(v)=w\text{.}\) Since \(v\neq 0\) we can extend \(\{v\}\) to a basis of \(V\text{,}\) say \(\{v,v_1,\ldots,v_{n-1}\}\text{.}\) For a set \(\{w,w_1,\ldots,w_{n-1}\}\subset V\) we define a map \(T(v)=w\) and \(T(v_i)=w_i\text{,}\) and we extend \(T\) linearly. Thus \(T\in\End_F(V)\) is a required map.
<Prev^TopNext>